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  1. #1
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    Smooth

    Let $\displaystyle f: \mathbb{R} \to \mathbb{R} $ be given. Assume that the square and cube of $\displaystyle f $ are smooth. Is $\displaystyle f $ smooth? That is , if $\displaystyle f \cdot f \in C^{\infty} $ and $\displaystyle f \cdot f \cdot f \in C^{\infty} $, does it follow that $\displaystyle f \in C^{\infty} $?
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  2. #2
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    Or maybe we can put it this way: if $\displaystyle f^{n} \in C^{\infty}$ and $\displaystyle f^{n-1} \in C^{\infty} $ does it follow that $\displaystyle f \in C^{\infty} $?
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  3. #3
    Super Member Rebesques's Avatar
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    Or maybe we can put it this way: if
    $\displaystyle
    f^{n} \in C^{\infty}
    $ and $\displaystyle
    f^{n-1} \in C^{\infty}
    $ does it follow that $\displaystyle
    f \in C^{\infty}
    $ ?

    Not in general. You need the extra assumption that $\displaystyle f^n,f^{n-1}$ are never zero.

    The proof is easy: Choose any natural number $\displaystyle m$. Since $\displaystyle gcd(n,n-1)=1,$ there are integers $\displaystyle q, \ p$ such that $\displaystyle pn+q(n-1)=1$. So $\displaystyle m=pmn+qm(n-1)$; Thus $\displaystyle f^m=(f^n)^{mp}(f^{n-1})^{qm}$ is differentiable as a product of differentiable functions. Note that the exponents $\displaystyle mp, \ mq$ may be negative, and so the assumption $\displaystyle f^n, f^{n-1}\neq0$ is essential.
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    I have another way to show this.

    If $\displaystyle f^3\not = 0$ for any point and $\displaystyle f^3\in C^{\infty}$ then $\displaystyle (f^3)^{1/3} \in C^{\infty}$ thus $\displaystyle f\in C^{\infty}$. Because the composition of infinitely differenciable functions is infinitely differenciable. Note, that $\displaystyle x^{1/3}$ is not differenciable at zero, which is why we need the condition that the function is never zero. However, I cannot come up with a conterexample.


    This does not work with $\displaystyle f^2$.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rebesques View Post
    Not in general. You need the extra assumption that $\displaystyle f^n,f^{n-1}$ are never zero.

    The proof is easy: Choose any natural number $\displaystyle m$. Since $\displaystyle gcd(n,n-1)=1,$ there are integers $\displaystyle q, \ p$ such that $\displaystyle pn+q(n-1)=1$. So $\displaystyle m=pmn+qm(n-1)$; Thus $\displaystyle f^m=(f^n)^{mp}(f^{n-1})^{qm}$ is differentiable as a product of differentiable functions. Note that the exponents $\displaystyle mp, \ mq$ may be negative, and so the assumption $\displaystyle f^n, f^{n-1}\neq0$ is essential.
    Ah! So my thought that we may construct the function f as $\displaystyle \frac{f^3}{f^2}$ is at least not out of the ballpark. (This, of course, gives the same limitation that $\displaystyle f \neq 0$ at any point in its domain.) But this can't be the whole story because if we have
    $\displaystyle f^2(x) = (x - 1)^4$
    and
    $\displaystyle f^3(x) = (x - 1)^6$

    the function $\displaystyle f(x) = (x - 1)^2$ is, in fact, $\displaystyle C^{\infty}$.

    How might we get around this? (Or am I barking up the wrong tree again? )

    -Dan
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  6. #6
    Super Member Rebesques's Avatar
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    Topsq, you have chosen a polynomial

    Powers of that are always differentiable!



    ps. What? 300 posts!! The Hacker must be feeling my breath on his back... :P
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rebesques View Post
    Topsq, you have chosen a polynomial

    Powers of that are always differentiable!



    ps. What? 300 posts!! The Hacker must be feeling my breath on his back... :P
    But my $\displaystyle f^3$ and $\displaystyle f^2$ are $\displaystyle C^{\infty}$ and there is a point in the domain of both that are 0. According to you and TPH my f should not be $\displaystyle C^{\infty}$. Or should we all be saying that we can't guarantee that such a $\displaystyle C^{\infty}$ f is possible under these conditions?

    -Dan
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  8. #8
    Super Member Rebesques's Avatar
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    According to you and TPH my f should not be...

    Alright you win, add "polynomials and no-denominator ever functions out"
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